a. V = a*t = 2 * 6 = 12 m/s.
V = 0.012km/s * 3600s/h = 43.2 km/h.
b. d1 = 0.5a*t^2 = 0.5*2*6^2 = 36 m.
d2 = V*t = 12 * 30 = 360 m.
V = Vo + a*t = 0.
12 + a*6 = 0, a = -2 m/s^2
d3 = Vo*t + 0.5a*t^2.
d3 = 12*6 - 1*5^2 = 47 m.
d = d1+d2+d3 = Total distance.
a car starts from rest and is accelerated uniformly and the rest of 2m/s2 for 6seconds it then maintaining a constant speed for half a minutes.The brake are then applied and then vehicles unifomly restarted to rest in 5seconds.find maxmum speed in km/h and the total distance covered in metres
3 answers
Correction: After the brakes were applied, V = Vo + a*t = 0.
12 + a*5 = 0, a = -2.4 m/s^2.
d3 = Vo*t + 0.5a*t^2.
d3 = 12*5 - 1.2*5^2 = 30 m.
d = d1+d2+d3 = 36 + 360 + 30 = 426 m. = Total distance covered.
12 + a*5 = 0, a = -2.4 m/s^2.
d3 = Vo*t + 0.5a*t^2.
d3 = 12*5 - 1.2*5^2 = 30 m.
d = d1+d2+d3 = 36 + 360 + 30 = 426 m. = Total distance covered.
B. Area of triangle 1(d1) = 1/2 × 6 × 12 = 200
Area of rectangle ( d2) = 30 × 12 = 360
Area of triangle 2 (d3) = 1/2 × 5 × 12 = 30
Total distance (d) = (d1 + d2 + d3)m
= (200 + 360 + 30)m
= 426m
A.Maximum speed (V) = a × t
= 2m/s^2 × 6s
= 12m/s
= 0.012km/s × 3600s/h
V = 43.2km/h
Area of rectangle ( d2) = 30 × 12 = 360
Area of triangle 2 (d3) = 1/2 × 5 × 12 = 30
Total distance (d) = (d1 + d2 + d3)m
= (200 + 360 + 30)m
= 426m
A.Maximum speed (V) = a × t
= 2m/s^2 × 6s
= 12m/s
= 0.012km/s × 3600s/h
V = 43.2km/h
1 answer